Thermodynamics & Energy Conservation?
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I'm having trouble understanding the energy conservation of thermodynamics. I understand that the first law of Thermodynamics is change in U = change in W + change in T, but when I actually need to apply energy conservation to a problem I run into trouble. The thing i'm having a hard time understanding is the change in U. I am used to thinking of U as the potential energy of a system, but here it means internal energy. what is the difference, exactly? For example, here is a conceptual problem: I fire a bullet of certain mass, M1, with certain velocity V1, through a tank containing certain mass of water, M2, and the bullet emerges with certain velocity, V2. How do I find the maximum temperature increase using energy conservation?
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Answer:
Look. With the first law of thermodynamics, it is NOTHING more than conservation of energy. Just account for ALL types of energy and you will be fine. Sometimes, you see the equation Q = ∆U + W. This means heat added to the system = change internal energy of the system + work done by the system. I don't know where the "change in T" came from, because change in temperature is NOT part of the first law of thermodynamics. Temperature is nothing more than a thermodynamic condition which tends to increase with internal energy per unit mass of a system. Why is heat added to the system positive and work done by the system positive? Heat is positive when entering and work is positive when leaving? Why? Two reasons: 1. It fits the calculus Pressure-volume formula for work with elegance (W = ∫P dV). 2. Desire of operation of the first thermodynamic cycles (heat engines). Historical reason: http://img63.imageshack.us/img63/3590/energysignconvention.jpg U does not mean potential energy in this context. U means INTERNAL ENERGY. The amount of "heat energy" contained in the system. Heat (Q) is different than internal energy, because internal energy is energy within the system, and heat is an exchange of thermal energy in or out of the system. From now on, call all potential energy with three letters, such as GPE (gravitational), SPE (strain as in springs and structural members), and EPE (electrical). Get that U out of your head for potential energy, and specify the type of internal energy as precisely as possible. ---------------- Bullet problem (assuming the tank of water is constrained). Don't resort to the standard first law formula, just think conservation of energy. Before the process occurred, the bullet had its initial kinetic energy KEi. Before the process occurred, the bullet had its final kinetic energy KEf, and the tank of water rose in internal energy ∆U. Thus, conservation of energy is: KEi = KEf + ∆U Solve for the least known: ∆U = KEi - KEf Express bullet's kinetic energies in terms of mass and velocity: KEi = 1/2*M1*V1^2 KEf = 1/2*M2*V2^2 Thus: ∆U = 1/2*M1*(V1^2 - V2^2) What does the change in internal energy manifest as in the water? How does it show evidence? A rise in temperature: ∆U = M2*c*∆T Thus: M2*c*∆T = 1/2*M1*(V1^2 - V2^2) Solve for ∆T: ∆T = M1*(V1^2 - V2^2)/(2*M2*c) Unit recommendations: Masses in kilograms Velocity in meters/second Specific heat capacity in Joules/kg-Celsius (or Joules/kg-Kelvin) Temperature in Celsius (or Kelvin). Kelvin is a better unit to choose as default, although in this situation it doesn't matter because by definition, Kelvin and Celsius have the same "size of degrees", and we are only talking about change in temperature. If I were your thermodynamics teacher, I would have you crunch numerous formulas containing temperature both blindly with each temperature scale, and have you conclude when it is important to use absolute temperature.
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Other answers
Look. With the first law of thermodynamics, it is NOTHING more than conservation of energy. Just account for ALL types of energy and you will be fine. Sometimes, you see the equation Q = ∆U + W. This means heat added to the system = change internal energy of the system + work done by the system. I don't know where the "change in T" came from, because change in temperature is NOT part of the first law of thermodynamics. Temperature is nothing more than a thermodynamic condition which tends to increase with internal energy per unit mass of a system. Why is heat added to the system positive and work done by the system positive? Heat is positive when entering and work is positive when leaving? Why? Two reasons: 1. It fits the calculus Pressure-volume formula for work with elegance (W = ∫P dV). 2. Desire of operation of the first thermodynamic cycles (heat engines). Historical reason: http://img63.imageshack.us/img63/3590/energysignconvention.jpg U does not mean potential energy in this context. U means INTERNAL ENERGY. The amount of "heat energy" contained in the system. Heat (Q) is different than internal energy, because internal energy is energy within the system, and heat is an exchange of thermal energy in or out of the system. From now on, call all potential energy with three letters, such as GPE (gravitational), SPE (strain as in springs and structural members), and EPE (electrical). Get that U out of your head for potential energy, and specify the type of internal energy as precisely as possible. ---------------- Bullet problem (assuming the tank of water is constrained). Don't resort to the standard first law formula, just think conservation of energy. Before the process occurred, the bullet had its initial kinetic energy KEi. Before the process occurred, the bullet had its final kinetic energy KEf, and the tank of water rose in internal energy ∆U. Thus, conservation of energy is: KEi = KEf + ∆U Solve for the least known: ∆U = KEi - KEf Express bullet's kinetic energies in terms of mass and velocity: KEi = 1/2*M1*V1^2 KEf = 1/2*M2*V2^2 Thus: ∆U = 1/2*M1*(V1^2 - V2^2) What does the change in internal energy manifest as in the water? How does it show evidence? A rise in temperature: ∆U = M2*c*∆T Thus: M2*c*∆T = 1/2*M1*(V1^2 - V2^2) Solve for ∆T: ∆T = M1*(V1^2 - V2^2)/(2*M2*c) Unit recommendations: Masses in kilograms Velocity in meters/second Specific heat capacity in Joules/kg-Celsius (or Joules/kg-Kelvin) Temperature in Celsius (or Kelvin). Kelvin is a better unit to choose as default, although in this situation it doesn't matter because by definition, Kelvin and Celsius have the same "size of degrees", and we are only talking about change in temperature. If I were your thermodynamics teacher, I would have you crunch numerous formulas containing temperature both blindly with each temperature scale, and have you conclude when it is important to use absolute temperature.
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